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Advanced Finite Element Analysis Prof. R. KrishnaKumar

Department of Mechanical Engineering Indian Institute of Technology, Madras

Lecture - 9

In the last class, we were looking at incompressibility.

(Refer Slide Time: 1:01)

Before that, there was a question on what we mean by dot? Say, sigma dot, epsilon dot or

lambda dot, what this dot really means? There is a small confusion. I had said that, said

this before and I am just repeating that; we are not into what is called as rate dependent

phenomena when we talk about plasticity. These rates are pseudo rates. So, sigma dot] is

perfectly equivalent to writing delta sigma, both of them are the same. That is in other

words, delta sigma when divided just by delta t, what is this? Time; time has nothing to

do with your application of load or loading time or whatever it is, it has nothing to do

with that. Just it is a pseudo time to only capture that the loading has to be applied in an

incremental fashion. That is all. So, this lambda dot, sigma dot, epsilon dot in one sense

conveys that the approach is incremental in nature. But we are going to see, there are

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some other things that come into picture later in the course, but right now you can say

that sigma dot and delta sigma, they are, both of them are equal.

So, the loading carries itself in time. In one sense the rate of loading is what manifests as

or comes out as sigma dot and so on. But, this rate of loading does not mean that we are

taking into account the inertial effects or rate dependent effects or strain rate effects and

so on. Once you are looking at strain rate effects, then the time should be the correct time.

It cannot be pseudo time. Is that clear? We will, we will give more explanation to it as we

go along in the course, but let us now look at a very important thing called

incompressibility.

Now, what is incompressibility? There was a question again in the last class, when we

finished that what is incompressibility? Incompressibility simply means that there is no

volume change. Incompressibility does not mean that we cannot compress it. We can

compress it, but when you compress it, the compressive strains that are developed are in

such a fashion that there is no change in volume. Is that clear? That is what it means.


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Now, we know that when it is compressible, we will prove it again in a more rigorous

setting later in the course that when incompressibility is the condition then nu equal to

0.5. Now, let us see, look at the difficulties that happens that take place when we have nu

is equal to 0.5. Since this is a material behavior, the culprit is going to be the D matrix.

All of you know that this is the elastic matrix. Now, let us split this D matrix into a part

which is going to be my problem or problematic part and a part which is not going to be

so very problematic part.


In order to do that let me define the shear modulus mu or G to be E divided by 2 into 1

minus nu and the bulk modulus B to be E divided by 3 into 1 minus 2 nu and say that this

D now can be split into two parts, one which goes along with the shear modulus, the

other part which goes along with the bulk modulus. Let us see how we can split this D.

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This is regular D, we are not talking about dt now; just we are going to look at regular D.

This D can now be written as, say for example, G into 4 by 3 minus 2 by 3 minus 2 by 3

minus 0 0 0, minus 2 by 3 4 by 3 minus 2 by 3 0 0 0 and so on, as one part plus the other

part consists of B into, B is my bulk modulus into and so on. Of course, the last rows are

zeros. Then it is, this becomes 0 0. Yeah, this part, this is zero; this is, no, third is minus 2

by 3 minus 2 by 3 0 0 0; the rest of it, 0 4 by 3 or sorry 1 1 1 and so on. Yeah, this is 6 by

6 matrix; this is a 6 by 6 matrix which of course we are talking about.

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This is the relationship between sigma is equal to D epsilon. So, we are splitting this D

into two parts, what goes into B transpose dB? Now, if you look at these two parts, the

culprit obviously is here, because you have 1 minus 2 nu term here. Now substituting, let

me call this as say DS, first part as DS and second part say as DB to indicate that one part

is formed by shear modulus and the other part formed by bulk modulus, so that my total

K consists of B transpose, instead of D, I am going to say that D shear modulus and bulk

modulus B d omega.

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This is DS and DB is what I am going to write, DS. So, I have two terms now in K. So, let

me call that then B transpose DS B d omega plus B transpose DB B d omega. So, in other

words, my K can be split into two parts say KS plus KB. Is it clear? Of course, DS is

complete thing. So, this B goes here. You have a B here, I mean of course it goes into DB.

What happens when we get nu is equal to 0.5? B goes to infinity. No, no; please note that

in G, I have only nu. So, when nu is equal to 0.5, this G does not go to infinity, but when

I have B which is there in this KB, second part where I have a B, then you will see that E

by 1 minus 2 nu, nu is equal to 0.5, so this goes to infinity.

As I approach nu to be 0.5, all my trouble starts; even at 0.49 or 95, 499 and so on, this

will make this K matrix, the situation will make the K matrix to be highly ill conditioned

and inverse here of this matrix is going to be difficult, because once it becomes ill

conditioned that means that it is going to a situation where it is, where it is almost you

know infinity, then I will have problems. At 0.5, completely incompressible, then it is

totally out. May be you can work up to 0.45, 44; it may not be a problem. Now what do I

do? What do I do to alleviate this problem?

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In a crude sense, I mean this is not, I am not going to talk total theory; we will come to

that a bit later. In a crude sense you may think that there is a solution. What is the

solution? Solution comes from a possible mistake I may make, use the integration rule.


Probably you would remember or studied, what is called as rank deficiency or hour

glassing. In our earlier class, we said that there is a particular integration rule I have to

use in order that I can integrate this accurately; polynomial of 2n minus 1 integrated

accurately with an order of n. In other words, we said that for a quadrilateral, I have to

use a 2 by 2 integration. What happens when I reduce this integration, when I go to 1 by

1 rule? I will get into what is called as rank deficiency, what is called as hour glassing.

The rank of the matrix is not sufficient enough for me to invert and make it stable. I will

get into singularity problems and so on. So, there is what is called rank deficiency when I

reduce the numerical integration and so I avoided it. So, I will go for full integration. If it

all I do reduced integration, I do other procedures called stabilization procedures. So, that

is called as hour glassing and in order to avoid hour glassing, I did some what is called as

stabilization procedure or in other words, in other words, you call this as hour glass

control.

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In fact, if you had used many of the explicit finite element code, you would see that most

of them use reduced integration and they will have a proper hour glass control. In other

words, by making a small, if I may call it a mistake, mistake within quotes, I got into

some trouble and I got out also by doing some damage exercises. So, major thing is that

that matrix is becoming singular when I did those things. Now, what I am going to do

now is very similar thing; I am going to make the mistake wantonly. How I am going to

do that? Having known that this kind of problems will exist, what I will do is simply take

this and perform an under integration or reduced integration.

In other words, I selectively integrate a part of the K matrix in a reduced fashion. So,

instead of 2 by 2, I may use 1; 1 by 1 row. If I need 3 by 3, I will use 2 by 2 row; I can

reduce the integration, order of integration. So, once I make that decision my problem to

a great extent is solved, because this KB matrix will become now singular. This will not

have, so this, the whole, though I will still have K from here. So, this contribution, this,

this problem of inverting this and going to infinity will not happen. That is called as

selective reduced integration. This problem of inverting this, I will not have difficulty. I

will have difficulty in the sense that that would be automatically taken care of by reduced

integration.

Now, this looks like a crime. People call this as crimes and so on; variational crimes,

common word that is used many times. This looks like a crime, but a beautiful paper by

Semo and Hughes showed that after all what you are doing is very well within the frame

work of variational approach. So, though it looks as if we can do the whole thing by

learning from a mistake, it interestingly has a very good theoretical background as well. It

comes from some of the variational principles which you might not have heard and we

will see that in this course. So, by doing this selective reduced integration, I can in one

word alleviate many of the problems. Is that clear?

How do you find out the order? But, you know the full order of integration that is

required, this depends upon the powers or the the polynomials that are there. That is why

I said 2n minus 1 polynomial is integrated exactly by end point integration rule. So,

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depending upon what is that you have, you can find out the exact and then you can reduce

it accordingly. Now, this is fine as far as this kind of simple elastic problem are

concerned and the approach seems to be sort of though not very correct, looks logical.

Now, how do we do a problem like plasticity or some other aspect where I cannot write

this very clearly like DS and DB and so on. Now, various methods have been proposed

and one of the methods which is very popular is called as B-bar method of Hughes. We

will see more theory after we do this. We are going to do the same selective reduced

integration, extend this concept what we have given here. I have not introduced, there is

lot more again simple things that you can talk about this. You will see that this B here

works like a penalty constant, but since we have not done lot more on penalty approach

and all this, I am not commenting on that, but suffice it for me to take me to the next

group of algorithms which come under what are called as B-bar method.


What essentially B-bar method does is to now alter not this D, but to alter B. After all you

can see that the culprit seems to be in the volumetric part and so can I now split B into

two parts - a deviatoric and a volumetric part and do something to the volumetric part of

K, which results from splitting this B, that B volume. Let us proceed instead of D

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splitting B. Yes, what is this B? Correct, the B is the strain displacement relationship; B

is the strain displacement relationship which is epsilon.


In this case epsilon happens to be B into the degrees of freedom, degrees of freedom,

which you can call this as u hat. Now, there is a difference between u and u hat. Note this

subtle difference between them. u is the displacement of a continuum. Now I am going to

use this because, I need some other, you know, alleviate to define these two separately.

So, u is a displacement of a continuum. In other words, u is a vector which can be written

as say this u is u x comma y and v x comma y. This u hat is the u at the say, nodal

locations; it is at the nodal locations. Suppose I have 1 2 3 4, then u1 u2 v1, u1 u2 u3 u4 v1

v2 v3 v4 that come under u hat. So, B is of course the strain displacement matrix. See, this

is the continuum u. This u and v are the continuum u. This u hat is after this continuum is

discretized and suppose I call this as 1 2 3 and 4, then u hat gives me a vector of the form

say u1 v1 u2 sorry v2 and so on up to u4 v4. This vector is what I call by this u hat.

Yeah, without hat; yes, that is because in the next formulation, I am going to use certain

things with u as well. So, that is why we are looking at this more closely. As you raise

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your level of understanding, you have to distinguish between u and u hat or in other

words continuum u and discretized u.


Say, may be in earlier classes we had put this as B u and understood that this u is this. We

never worried about the continuum u. Sometimes it may be necessary to have that also,

so that strain epsilon we will define it as Su. S is my operator. We will define that again

bit later. So, let us now define it like this. Now of course, B is a matrix. B is a matrix

which consists of say B1 B2 B3 and so on up to B into number of elements.

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So suppose this is, what is it? You know what is the size? This is say 6; 6 by 1 and this

happens to be, is it 6 by 1? Yes, for we are talking about this quad, 8 by 1 and so 6 by 8.

Is that clear?


So, this B matrix has a sub matrix as well with two in each can be written as B1, B1 is

itself is a sub matrix, 0 0, 0 B2 0, 0 0 B3, all of them are matrices; 0 B2 or B1 B2. So,

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depending upon whether you write epsilon23 or epsilon13 and then B3 0 B1 and B2 B1 B3

B2 or B3, yeah B2 B3. Let us check that out. So, suppose I have epsilon11, this will

become, so that should be B1 0, sorry, it should be B1 B2 yes, just we stick on to the same

thing ….. sorry B2 B1; B2 B1 0 B3 B2 B3 B1. So, let us put it like this. Anyway you know

what this B’s are.

Now, what we are going to do essentially is to split this B into two parts, a dilatational

part and a deviatoric part.


Let me call this B dilatational part to be say, one-third of, this comes out very easily; B1,

B1 comes from that, this one B1 B2 B3, B1 B2 B3, B1 B2 B3, 0 0 0, 0 0 0, 0 0 0. Now, how

did I get that? One minute, how did I get that? I got this from the fact that the dilatational

part is given by epsilon11 plus epsilon22 plus epsilon33 and similar to what we did for

pressure by 3, so, epsilon 11 22 33 consists of sum of B1 and corresponding to u1, B2

corresponding to u2, B3 corresponding to u3, so, B1 B2 B3, together forms epsilon 11 22

33. So, that is why we had that by 3 and so that gives the dilatational part.

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This B, let, let us, that, that was covered in the last class or last course, but it was, it is

quite simple. How did we get epsilon? You know, just look at how we derived epsilon.


So, epsilon, say epsilon11, what is epsilon11? Dow u, yeah, dow u by dow x; so, in other

words, dow u by dow x1 if you want to keep u as u1 u2 u3 and so on. So, you say that dow

u1 by dow x1; epsilon22 dow u2 by dow x2 and so on. So, what is epsilon12? Half of u i

comma j plus u j comma i. Dow u1 by dow x2 plus dow u2 by dow x1 and similarly you

can write. Usually like, I said in the last course we will not use epsilon, most of the time

we will use only gamma12, the engineering shear stress.

Yeah, 1 by 2 will not be there. That is why we have removed that. So, this actually can be

written in a matrix form. Let me remove this, we will come back to this in a minute; we

will be repeating some of the things.

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This will, can be written in the matrix form as epsilon is equal to u, with first entry being

dow by say dow x1. Suppose, I am only looking at a two dimensional case, you can

extend it very easily to three dimensional case, so that let me see that is epsilon11, next is

epsilon22, say gamma12; only these three I am going to see. Then, this can be written as

dow by dow x1 0 0, 0 dow by dow x2 0. Then I have to multiply, so, first one will be dow

by dow x2 dow by dow x1 0. This would be the, for these three. So, I will have actually,

because 33 was not there, this was all zero. In other words, you can look it at even as just

like that.

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Of course, when I want to have a complete set then, I may, I may have to extend this into

3 rows or rather 3 columns and 6 rows. So, this matrix is what I would call as a S matrix.

Note that this u is my continuum u. This is what now is replaced by N into discretized u.

That is what I called as u hat.


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So S into N into u. Now, let us not worry about the isoparametric; N is our shape

function, obviously so, u hat this is, this is what is called as my u hat and N u. Now, this

multiplied by this S N, this can be written as S N u. So, this S N is what is called as B,

sorry, u hat; that is u hat. So, S N is what is called as B into u hat. So, you can down write

B very easily. Is that clear? Yes; this is for, this for one u. So, if you, if I now put down

all the nodes correspondingly I will have in u, I will have corresponding to all the nodes,

so it will, I will have u1.


Let me call this as from u1 to u8 to make things clear again. So, u1 u2 u3 u4 u5 u6 u7 and u8.

So, u will be defined in terms of N1 N2 N3 N4. So, how do you write N now? N will be,

now will be 3. See, this will be say, let, let us come back to dimensions of this. Suppose I

am looking at 6; so, 6 here. So, 6 by 1 is equal to, S will be 3, 6 by 3. What will be N? N

will be 3 by 8, sorry and u will be 8 by 1, so that 6, 6 by 3 into 3 by 8 6 by 8 into 8 by 1.

Now, my B matrix I said is 6 by 8. That is what I wrote this as in a form which is, you

know, combining this I wrote it like that, B1 B2. That is why I said B1 is a sub matrix. Is

that clear?

So, symbolically, I mean I am, I am not expanding all this.

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Symbolically I am writing it so that you will, I mean you know here after I will not write

u1 u2, I will just say that this is the u vector. That is all. Fine; so please note that the

number of degrees of freedom also entered into this B matrix.


So, this B matrix is the one which I say can be split up into two parts. One is B

dilatational part plus B deviatoric part; B dilatational part and B deviatoric part. The

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dilatational part consists of one-third of B’s, one-third of say, the first part of B is or in

other words, the dilatational part comes from the fact that it is epsilon 11 plus 22 plus 33.

So, I can write this say, as B1 B2 B3; it is the same B1 B2 B3, B1 B2 B3. This is for, actually

what we did here note that this is for 2D case. If I have to actually, this a small, I think we

did not do this correctly; sorry, one minute.


We did not do that correctly, because we are looking at 3D case. So 3D case is, actually

we have extended it to 3D case. So, in 3D case this is equal to 6 by 1 and S is, there are

how many rows? 3; so how many, how many columns? 3 columns and 6 rows, so, 6 into

3. Then here it is 3 into, say for example, you are looking at a brick element now. Brick

element, 8; 8 into, yes, 8 into 2, number of degrees of freedom, 16. So this should be 3

into 16; so, 16 into 1. Which one? Yeah, 8 into 3, 24 right, not 62, not 2; 3, so, 3 into 24

and 24 into 1, 24 into 1; correct. So, 6 into 3, 3 into 24, 24 into 1, so that now B matrix is

actually S into N, which means 6 into 24; 6 into 24 and so each of this B1 are …….

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Yeah, so, that is why now B dilatational one-third is written as B1 B2 B3. These are

matrices, now you know what they are? B1 B2 B3, B1 B2 B3, 0 0 0, 0 0 0, 0 0 0 ; one-third.

Yes; wait a minute. Let me finish it. So, plus I have a deviatoric part such that the sum of

this deviatoric plus dilatational part will give me B. Is that clear?

No; I am splitting this B into a dilatational part in the same fashion that I can split strains,

because after all these are strain displacement relationship. So, in the same fashion I can

split strains I am splitting B into the dilatational part and deviatoric part. So, the

dilatational part is this and deviatoric part is the other one. So, in fact if I am going to

write down these two before we write down the equation and substitute this into my K

matrix, you will see that I can write that down as my K matrix or KT matrix, can be

written as integral omega say B deviatoric plus B dilatational.

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I am going to introduce lot more theory, but quick glance of what we are going to do,

transpose D into, same way B is again split B deviatoric plus B dilatational or DT into d

omega. So, wherever I get the dilatational part, I am going to under integrate and

deviatoric part is full integration. So, in other words, this B1 B2 what you see here will be

replaced by new B’s called B bars which is the result of under integration. There are

various ways of getting B bars we will see that in the next class. One is by under, under

integration and you will see that we can get the same formulation by going over to

another formulation called mixed formulation. It is possible to go to a formulation called

mixed formulation; I will just explain it in a minute and come back to this again. But, I

just want to say here that now I am splitting B.

The essence of the story is that I am splitting B. I am splitting this into a deviatoric part

and a dilatational part and I am concentrating on the dilatational part and under

integrating it and the deviatoric part I am doing full integration. In other words, if I just

look at the implementation, what will happen is that in an element I will have 1 2 3 4, 4

Gauss points for a 2D element, correspondingly 2 by 2 by 2 in a 3D element for not, it is

not a higher order element, it is a 4 noded element and these 4 nodes or 4 Gauss points

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rather are used to integrate the deviatoric part and I will use one Gauss point at the center

to integrate the dilatational part. So, this is called as selective reduced integration.

Yes. Yeah, usually you do not consider them, but you take that separately as deviatoric

DT. That is why we would call that as a B-bar. You will see how it comes up in the next

part of the derivation in the mixed formulation, but you will, usually you take deviatoric

part separately into DT, dilatational part separately. So, you will have deviatoric DT

deviatoric, dilatational DT dilatational. These are the two things that you will take them

up separately. In other words, it is not very correct to split that. In fact, I should have

written this as B consisting of, actually this comes from my P matrix, from the B

transpose sigma.


Actually I should have, I should not have written it like this. Actually we should have

written it as B transpose sigma d omega and from here I should have written this as B

deviatoric plus B dilatational sigma d omega and so on. In other words, you will have

only two parts B deviatoric d B deviatoric plus dilatational part. So, that is what you

under integrate. So, you will have, actually if you really look at it, you will have 5 Gauss

points.

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These four are the ones which are used for deviatoric part and the center Gauss point is

what is used for the dilatational part. So, you will have, in other words, the history

variables are stored in all these 5 Gauss points as well. Now, what is the theoretical

background? We will, we will do that in the next class to this. The theoretical background

for this comes from what is called as mixed formulation and going over to a variational

approach called as Hu-Washizu variational principle.

In the mixed formulation, in the mixed formulation you have instead of displacement

alone as a variable, you also have sometimes p as well as volumetric strain epsilonv.

When I have u p and v, epsilonv as a variable, then we call this as, there are three things,

so, these are your three mixed formulations or sometimes you will have only u and stress.

These are two point mixed formulation. We will talk about mixed formulation and Hu-

Washizu rule or variational formulation in the next class and tie that up and go into the

details of this B and how that comes about Hu-Washizu principle in the next class.

Yes, because in the plasticity problem, can we decompose DT clearly? It is not possible as

simple as what we did for an elastic case, because elastic case we had the shear modulus

and the bulk modulus clearly defined and quite easy to demarcate it. We will have

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problems when we do that in the plastic, in the plasticity part of it. So, let us look at the

mixed formulation and then talk about the rest of it.

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